You allocate $10$ identical lots across $4$ accounts, and an account may receive none. How many allocations are there?

The lots are identical and the accounts are distinct, which is the stars-and-bars setup. Lay out ten stars and choose where to place three bars to cut them into four groups, giving $\binom{10 + 4 - 1}{4 - 1} = \binom{13}{3} = \frac{13 \times 12 \times 11}{6} = 286$. Allowing an empty account is what makes it $\binom{13}{3}$ rather than $\binom{9}{3} = 84$. If the lots were distinguishable the answer would instead be $4^{10}$, which is a different question entirely.

Takeaway: $k$ identical items in $n$ distinct boxes is $\binom{n+k-1}{k}$.